HSC Mathematics Advanced, Extension 1, and Extension 2/3-Unit/HSC/Primitive of sin and cos squared

The integration of these functions can be performed by the substitution of $\sin ^{2}\alpha$ or $\cos ^{2}\alpha$ by a function in terms of $\cos 2\alpha$ respectively.

Deriving the substituted functions

By the double-angle formula, $\left.\cos 2\alpha =2\cos ^{2}\alpha -1\right.$. Rearranging to make $\cos ^{2}\alpha$ the subject, $\cos ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha$.

We know by the Pythagorean identity $\left.\sin ^{2}\theta +\cos ^{2}\theta =1\right.$ that $\left.\cos ^{2}\alpha =1-\sin ^{2}\alpha \right.$, so $1-\sin ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha$. Rearranging, $\sin ^{2}\alpha ={\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha$.

Finding the primitive

By substituting $\sin ^{2}\alpha \,$ for ${\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha$, $\int \sin ^{2}\alpha d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha d\alpha$. Similarly for $\cos ^{2}\alpha \,$,

$\int \cos ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha$

These can be integrated by using the primitives of cosine and sine

$\int \cos \alpha d\alpha =\sin \alpha +c$

$\int \sin \alpha d\alpha =-\cos \alpha +c$

For cos² α

$\int \cos ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha$

$\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c$

$\int \cos ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c$

For sin² α

$\int \sin ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha$

$\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c$

$\int \sin ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c$